试用
Spring-JDBC.我使用
this作为参考.我正在尝试获得一个名称相同的演员.运行此代码给了我所需的结果：
- public List<String> getActorsWithSameLastName(String lastName,NamedParameterJdbcTemplate template) {
- String query = "SELECT FIRSTNAME FROM ACTORS WHERE LASTNAME=:LASTNAME";
- Map<String,String> paramMap = new HashMap<String,String>();
- paramMap.put("LASTNAME",lastName);
- return template.queryForList(query,paramMap,String.class);
- }
我有一个列表< String>的姓氏我如何获得一个列表的演员与我的列表？我是否迭代姓氏列表,并且每次调用getActorsWithSameLastName(),或者Spring提供了一种迭代方式,并为我获取结果？请指教.
解决方法
使用IN条款..
How to use SELECT IN clause in JDBCTemplates?
- List<String> lastnames= new ArrayList();
- Map namedParameters = Collections.singletonMap("lastnamevalues",lastnames);
- StringBuffer recordQueryString = new StringBuffer();
- recordQueryString.append("select FIRSTNAME,LASTNAME from ACTORS where lastname in (:lastnamevalues) ");
- List nameInvolments = this.namedparameterJdbcTemplate.query(recordQueryString.toString(),namedParameters,new MyMapper());
